1.          Given that $\log 2.345=0.3701 .$ What are the characteristics and the mantissas of each of the followings?

(a) $\log 234,500$

(b) $\log 0.0002345$

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ \ \ \ \log 2.345=0.3701\\\\\text{(a)}\ \ \ \log 234,500\\\\\ \ \ =\log \left( {2.345\times {{{10}}^{5}}} \right)\\\\\ \ \ =\log {{10}^{5}}+\log 2.345\\\\\ \ \ =5+0.3701\\\\\ \ \ =5.3701\\\\\therefore \ \text{characteristic}=5\\\\\ \ \ \text{mantissa}=.3701\\\\\\\text{(b)}\ \ \ \log 0.0002345\\\\\ \ \ =\log \left( {2.345\times {{{10}}^{{-4}}}} \right)\\\\\ \ \ =\log {{10}^{{-4}}}+\log 2.345\\\\\ \ \ =-4+0.3701\\\\\therefore \ \text{characteristic}=-4\\\\\ \ \ \text{mantissa}=.3701\end{array}$

2.          Using $\log _{10} 2.74=0.4377, \log _{10} 2.83=0.4518, \log _{10} 5.97=0.7760$, $\log _{10} 6.21=0.7931, \log _{10} 8.18=0.9128$ and $\log _{10} 9.27=0.9671$, compute

(a) $\left(\displaystyle\frac{28.3}{597 \times 621}\right)^{2}$

(b) $\displaystyle\frac{274^{\frac{1}{3}}}{927 \times 818}$

(c) $\displaystyle\frac{28.3 \sqrt{0.621}}{597}$

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ {{\log }_{{10}}}2.74=0.4377,\\\ \ \ \ \ {{\log }_{{10}}}2.83=0.4518,\\\ \ \ \ \ {{\log }_{{10}}}5.97=0.7760,\\\ \ \ \ \ {{\log }_{{10}}}6.21=0.7931,\\\ \ \ \ \ {{\log }_{{10}}}8.18=0.9128,\\\ \ \ \ \ {{\log }_{{10}}}9.27=0.9671.\\\text{(a)}\ \ \text{Let}\ x={{\left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)}^{2}},\ \text{then}\\\ \ \ \ \ \log x=\log {{\left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)}^{2}}\\\ \ \ \ \ \ \ \ \ \ \ \ =2\log \left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log (597\times 621)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log 597-\log 621} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log 597-\log 621} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {1.4518-2.7760-2.7931} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =-8.2346\\\ \ \ \ \ x={{10}^{{-8.2346}}}\\\\\text{(b)}\ \ \text{Let}\ x=\displaystyle\frac{{{{{274}}^{{\displaystyle\frac{1}{3}}}}}}{{927\times 818}},\ \text{then}\\\ \ \ \ \ \log x=\log \left( {\displaystyle\frac{{{{{274}}^{{\displaystyle\frac{1}{3}}}}}}{{927\times 818}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\log 274-\log \left( {927\times 818} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\log 274-\left( {\log 927+\log 818} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\left( {2.4377} \right)-2.9671-2.9128\\\ \ \ \ \ \ \ \ \ \ \ \ =-5.0673\\\ \ \ \ \ x={{10}^{{-5.0673}}}\\\\\text{(c)}\ \ \text{Let}\ x=\displaystyle\frac{{28.3\sqrt{{0.621}}}}{{597}},\ \text{then}\\\ \ \ \ \ \log x=\log \left( {\displaystyle\frac{{28.3\sqrt{{0.621}}}}{{597}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\log 28.3+\displaystyle\frac{1}{2}\log 0.621-\log 597\\\ \ \ \ \ \ \ \ \ \ \ \ =1.4518+\displaystyle\frac{1}{2}\left( {-1+0.7931} \right)-2.7760\\\ \ \ \ \ \ \ \ \ \ \ \ =-1.4277\\\ \ \ \ \ x={{10}^{{-1.4277}}}\end{array}$

3.          Express each to a natural logarithm in terms of $\ln 2$ and $\ln 5$.

(a) $\log_{8}25$

(b) $\log_{5}64$

(c) $\log_{25}100$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ \ {{\log }_{8}}25\ \\\ \ \ \ =\displaystyle\frac{{\ln 25}}{{\ln 8}}\\\ \ \ \ =\displaystyle\frac{{\ln {{5}^{2}}}}{{\ln {{2}^{3}}}}\\\ \ \ \ =\displaystyle\frac{{2\ln 5}}{{3\ln 2}}\\\\\text{(b)}\ \ \ \ {{\log }_{5}}64\ \\\ \ \ \ =\displaystyle\frac{{\ln 64}}{{\ln 5}}\\\ \ \ \ =\displaystyle\frac{{\ln {{2}^{6}}}}{{\ln 5}}\\\ \ \ \ =\displaystyle\frac{{6\ln 5}}{{\ln 2}}\\\\\text{(c)}\ \ \ \ {{\log }_{{25}}}100\ \\\ \ \ \ =\displaystyle\frac{{\ln 100}}{{\ln 25}}\\\ \ \ \ =\displaystyle\frac{{\ln ({{5}^{2}}\times {{2}^{2}})}}{{\ln {{5}^{2}}}}\\\ \ \ \ =\displaystyle\frac{{2\ln 5+2\ln 2}}{{2\ln 5}}\\\ \ =1+\displaystyle\frac{{\ln 2}}{{\ln 5}}\ \ \ \end{array}$

4.          Write each expression as a single logarithm.

(a) $3\ln \left( {t + 5} \right) - 4\ln t - 2\ln \left( {s - 1} \right)$

(b) $2\ln x + 5\ln y - \displaystyle\frac{1}{2}\ln z$

(c) $\displaystyle \frac{1}{3}\ln a - 6\ln b + 2$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ \ \ \ 3\ln \left( {t+5} \right)-4\ln t-2\ln \left( {s-1} \right)\\\\ \ \ \ \ =\ln {{\left( {t+5} \right)}^{3}}-\ln {{t}^{4}}-\ln {{\left( {s-1} \right)}^{2}}\\\\ \ \ \ \ =\ln \displaystyle\frac{{{{{\left( {t+5} \right)}}^{3}}}}{{{{t}^{4}}{{{\left( {s-1} \right)}}^{2}}}}\\\\\\ \text{(b)}\ \ \ \ 2\ln x+5\ln y-\displaystyle\frac{1}{2}\ln z\\\\ \ \ \ \ =\ln {{x}^{2}}+\ln {{y}^{5}}-\ln \sqrt{z}\\\\ \ \ \ \ =\ln \displaystyle\frac{{{{x}^{2}}\cdot {{y}^{5}}}}{{\sqrt{z}}}\\\\\\ \text{(c)}\ \ \ \ \displaystyle\frac{1}{3}\ln a-6\ln b+2\\\\ \ \ \ \ =\ln {{a}^{{\frac{1}{3}}}}-\ln {{b}^{6}}+\ln {{e}^{2}}\\\\ \ \ \ \ =\ln \displaystyle\frac{{\sqrt[3]{a}\cdot {{e}^{2}}}}{{{{b}^{6}}}} \end{array}$

5.          Solve the following logarithmic equations for $x$.

(a) $\ln \left( x \right) + \ln \left( {x + 3} \right) = \ln \left( {20 - 5x} \right)$

(b) $2\ln \left( {\sqrt x } \right) - \ln \left( {1 - x} \right) = 2$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ln (x)+\ln \left( {x+3} \right)=\ln \left( {20-5x} \right)\\\ \\\ \ \ \ \ \ \ln \left( {x\left( {x+3} \right)} \right)=\ln \left( {20-5x} \right)\\\\\ \ \ \ \ \ x\left( {x+3} \right)=20-5x\\\\\ \ \ \ \ \ {{x}^{2}}+8x-20=0\\\\\ \ \ \ \ \ (x+10)(x-2)=0\\\\\ \ \ \ \ \ x=-10\ \text{or}\ x=2\\\\\ \ \ \ \ \ \text{Since }x>0,\ x=-10\ \text{is impossible}\text{.}\\\\\ \ \ \therefore \ \ x=2\\\\\ \ \ \ \ \\\text{(b)}\ \ \ 2\ln \left( {\sqrt{x}} \right)-\ln \left( {1-x} \right)=2\\\ \\\ \ \ \ \ \ln \left( {\displaystyle\frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}}}{{1-x}}} \right)=2\\\\\ \ \ \ \ \displaystyle\frac{x}{{1-x}}={{e}^{2}}\\\ \\\ \ \ \ \ x={{e}^{2}}-{{e}^{2}}x\\\\\ \ \ \ \ {{e}^{2}}x+x={{e}^{2}}\\\\\ \ \ \ \ x\left( {{{e}^{2}}+1} \right)={{e}^{2}}\\\\\ \ \ \ \ x=\displaystyle\frac{{{{e}^{2}}}}{{{{e}^{2}}+1}}\end{array}$

6.          Solve the equations:

(a) $x - xe^{5x + 2} = 0$

(b) $7 + 15e^{1 - 3x} = 10$ using $\ln 5 = 1.6094$.

(c) $4e^{1 + 3x} - 9e^{5 - 2x} = 0$ using $\ln 2=0.6931$ and $\ln 3=1.0986$.

No(3) မှ No(6) အထိ မေးခွန်းများသည် ပြဌာန်းစာအုပ်တွင် မပါဝင်ပါ Natural Logarithm ဆိုင်ရာ ပုစ္တာများကို တိုးချဲ့လေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးထားခြင်း ဖြစ်ပါသည်။

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ x-x{{e}^{{5x+2}}}=0\\\ \\\ \ \ \ \ \ x\left( {1-{{e}^{{5x+2}}}} \right)=0\\\\\ \ \ \ \ \ x=0\ \text{or}\ 1-{{e}^{{5x+2}}}=0\\\\\ \ \ \ \ \ x=0\ \text{or}\ {{e}^{{5x+2}}}=1\\\\\ \ \ \ \ \ x=0\ \text{or}\ 5x+2=\ln 1\\\\\ \ \ \ \ \ x=0\ \text{or}\ 5x+2=0\\\\\ \ \therefore \ \ \ x=0\ \text{or}\ x=-\displaystyle\frac{2}{5}\\\\\ \ \ \ \ \\\text{(b)}\ \ \ 7+15{{e}^{{1-3x}}}=10\\\ \\\ \ \ \ \ \ 15{{e}^{{1-3x}}}=3\\\\\ \ \ \ \ \ {{e}^{{1-3x}}}=\displaystyle\frac{1}{5}\\\ \\\ \ \ \ \ \ 1-3x=\ln \left( {\displaystyle\frac{1}{5}} \right)\\\\\ \ \ \ \ \ 1-3x=\ln \left( {{{5}^{{-1}}}} \right)\\\\\ \ \ \ \ \ 1-3x=-\ln \left( 5 \right)\\\\\ \ \ \ \ \ 3x=1+\ln \left( 5 \right)\\\\\ \ \ \ \ \ 3x=1+\text{1}\text{.6094}\\\\\ \ \ \ \ \ 3x=2.\text{6094}\\\\\ \ \ \ \ \ x=\text{0}\text{.8698}\\\\\\\text{(c)}\ \ \ 4{{e}^{{1+3x}}}-9{{e}^{{5-2x}}}=0\\\\\ \ \ \ \ 4{{e}^{{1+3x}}}=9{{e}^{{5-2x}}}\\\\\ \ \ \ \ \displaystyle\frac{{{{e}^{{1+3x}}}}}{{{{e}^{{5-2x}}}}}=\displaystyle\frac{9}{4}\\\\\ \ \ \ \ {{e}^{{5x-4}}}=\displaystyle\frac{9}{4}\\\\\ \ \ \ \ 5x-4=\ln \displaystyle\frac{{{{3}^{2}}}}{{{{2}^{2}}}}\\\\\ \ \ \ \ 5x=2\left( {\ln 3-\ln 2} \right)+4\\\\\ \ \ \ \ x=\displaystyle\frac{2}{5}\left( {\ln 3-\ln 2+2} \right)\\\\\ \ \ \ \ x=\displaystyle\frac{2}{5}\left( {\text{1}\text{.0986}-\text{0}\text{.6931+2}} \right)\\\\\ \ \ \ \ x=\text{0}\text{.9622}\end{array}$

1.          If $\log _{a} b+\log _{b} a^{2}=3,$ find $b$ in terms of $a$.

Show/Hide Solution

$\begin{array}{l}{{\log }_{a}}\,b+{{\log }_{b}}\,{{a}^{2}}=3\\\\{{\log }_{a}}\,b+2{{\log }_{b}}a=3\\\\{{\log }_{a}}\,b+\displaystyle\frac{2}{{{{{\log }}_{a}}\,b}}=3\\\\\text{Let}\ {{\log }_{a}}\,b=u,\ \text{then}\\\\u+\displaystyle\frac{2}{u}=3\\\\\therefore \ \ {{u}^{2}}+2=3u\\\\\therefore \ \ {{u}^{2}}-3u+2=0\\\\\therefore \ \ (u-1)(u-2)=0\\\\\therefore \ \ {{\log }_{a}}\,b=1\ \text{or}\ {{\log }_{a}}\,b=2\\\\\therefore \ \ \,b=a\ \text{or}\ \,b={{a}^{2}}\end{array}$

2.          Show that

(a) $\log _{4} x=2 \log _{16} x$.

(b) $\log _{b} x=3 \log _{b^{3}} x \quad$.

(c) $\log _{2} x=\left(1+\log _{2} 3\right) \log _{6} x$.

Show/Hide Solution

$\begin{array}{l}(\text{a})\ \ \ \text{LHS}={{\log }_{4}}x\\\\\ \ \ \ \ \ \text{RHS}=2{{\log }_{{16}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\log }_{{{{4}^{2}}}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{2}{2}{{\log }_{4}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{4}}x\\\\\ \ \therefore \ \ {{\log }_{4}}x=2{{\log }_{{16}}}x\\\\\\(\text{b})\ \ \text{LHS}={{\log }_{b}}x\\\\\ \ \ \ \ \text{RHS}=3{{\log }_{{{{b}^{3}}}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{3}{3}{{\log }_{b}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}x\\\\\ \ \therefore \ \ {{\log }_{b}}x=3{{\log }_{{{{b}^{3}}}}}x\\\\\\(\text{c})\ \ \text{LHS}={{\log }_{2}}x\\\\\ \ \ \ \ \text{RHS}=\left( {1+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {{{{\log }}_{2}}2+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{2}}6\cdot {{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{{{{{\log }}_{6}}2}}\cdot {{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{{{{{\log }}_{6}}x}}{{{{{\log }}_{6}}2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{2}}x\\\\\ \ \therefore \ \ {{\log }_{2}}x=\left( {1+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\end{array}$

3.          If $a=\log _{b} c, b=\log _{c} a$ and $c=\log _{a} b,$ prove that $a b c=1$.

Show/Hide Solution

$ \begin{array}{l}\ \ \ a={{\log }_{b}}c,\ b={{\log }_{c}}a,\ c={{\log }_{a}}b\\\\\therefore \ \ abc={{\log }_{b}}c\cdot \ {{\log }_{c}}a\cdot {{\log }_{a}}b\\\\\therefore \ \ abc=\displaystyle\frac{{\log c}}{{\log b}}\cdot \ \displaystyle\frac{{\log a}}{{\log c}}\cdot \displaystyle\frac{{\log b}}{{\log a}}\ \left[ {\because \ \text{by L8}} \right]\\\\\therefore \ \ abc=1\end{array}$

4.          Show that

(a) $\left(\log _{10} 4-\log _{10} 2\right) \log _{2} 10=1$.

(b) $2 \log _{2} 3\left(\log _{9} 2+\log _{9} 4\right)=3$.

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ \left( {{{{\log }}_{{10}}}4-{{{\log }}_{{10}}}2} \right)\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}\displaystyle\frac{4}{2}\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}2\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}2\cdot \displaystyle\frac{1}{{{{{\log }}_{{10}}}2}}\\\\\ \ \ =\ 1\\\\\\\text{(b)}\ \ \ 2\cdot {{\log }_{2}}3\cdot \left( {{{{\log }}_{9}}2+{{{\log }}_{9}}4} \right)\\\\\ \ \ ={{\log }_{2}}{{3}^{2}}\cdot {{\log }_{9}}(2\times 4)\\\\\ \ \ ={{\log }_{2}}9\cdot {{\log }_{9}}8\\\\\ \ \ ={{\log }_{2}}9\cdot {{\log }_{9}}{{2}^{3}}\\\\\ \ \ ={{\log }_{2}}9\cdot 3{{\log }_{9}}2\\\\\ \ \ =3\cdot \displaystyle\frac{1}{{{{{\log }}_{9}}2}}\cdot {{\log }_{9}}2\\\\\ \ \ =3\end{array}$

5.          Compute

(a) $3^{\log _{2} 5}-5^{\log _{2} 3}$

(b) $4^{\log _{2} 3}$

(c) $2^{\log _{2 \sqrt{2}} 27}$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ {{3}^{{{{{\log }}_{2}}\,5}}}-{{5}^{{{{{\log }}_{2}}\,3}}}\\\\\ \ \ \ ={{3}^{{{{{\log }}_{2}}\,5}}}-{{3}^{{{{{\log }}_{2}}\,5}}}\\\\\ \ \ \ =0\\\\\text{(b)}\ \ \ {{4}^{{{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{2}}} \right)}^{{{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{{{{{\log }}_{2}}3}}}} \right)}^{2}}\\\\\ \ \ \ ={{3}^{2}}\\\\\ \ \ \ =9\\\\\\\text{(c)}\ \ \ {{2}^{{{{{\log }}_{{2\sqrt{2}}}}27}}}\\\\\ \ \ \ ={{2}^{{{{{\log }}_{{{{2}^{{3/2}}}}}}{{3}^{3}}}}}\\\\\ \ \ \ ={{2}^{{\frac{2}{3}\times 3{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{2}^{{2{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{{{{{\log }}_{2}}3}}}} \right)}^{2}}\\\\\ \ \ \ ={{3}^{2}}\\\\\ \ \ \ =9\end{array}$

6.          Using properties of logarithm, solve the following equations.

(a) $ {{\log }_{x}}10+{{\log }_{{{{x}^{2}}}}}10=2$

(b) $25^{\log _{10} x}=5+4 x^{\log _{10} 5}$

(c) $9^{\log _{3}\left(\log _{2} x\right)}=\log _{2} x-\left(\log _{2} x\right)^{2}+1$

(d) $\log _{2} x+\log _{4} x+\log _{16} x=7$

(e) $\log _{7} 2+\log _{49} x=\log _{\frac{1}{7}} \sqrt{3}$

(f) $\log _{3} x-\log _{\frac{1}{3}} x^{2}=6$

(g) $ {{\log }_{x}}(9{{x}^{2}})\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4$

No (6) သည် ပြဌာန်းစာအုပ်တွင် မပါဝင်ပါ။ Change of Base နှင့် သက်ဆိုင်သော ဉာဏ်စမ်းမေးခွန်း များကို ထပ်ဆင့်လေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးထားခြင်း ဖြစ်ပါသည်။

Show/Hide Solution

$ \begin{array}{l} \text{(a)}\ \ {{\log }_{x}}10+{{\log }_{{{{x}^{2}}}}}10=2\\\\ \ \ \ \ \ {{\log }_{x}}10+\displaystyle\frac{1}{2}{{\log }_{x}}10=2\\\\ \ \ \ \ \ \displaystyle\frac{3}{2}{{\log }_{x}}10=2\\\\ \ \ \ \ \ {{\log }_{x}}10=\displaystyle\frac{4}{3}\\\\ \ \ \ \ \ {{x}^{{\frac{4}{3}}}}=10\\\\ \ \ \ \ \ x={{10}^{{\frac{3}{4}}}}\\\\\\ \text{(b)}\ \ {{25}^{{{{{\log }}_{{10}}}x}}}=5+4{{x}^{{{{{\log }}_{{10}}}5}}}\\\\ \ \ \ \ \ {{\left( {{{5}^{2}}} \right)}^{{{{{\log }}_{{10}}}x}}}=5+4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}\\\\ \ \ \ \ \ {{5}^{{2\cdot }}}^{{{{{\log }}_{{10}}}x}}=5+4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}\\\\ \ \ \ \ \ {{\left( {{{5}^{{{{{\log }}_{{10}}}x}}}} \right)}^{2}}-4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}-5=0\\\\ \ \ \ \ \ \text{Let}\ {{5}^{{{{{\log }}_{{10}}}x}}}=u,\ \text{then}\\\\ \ \ \ \ \ {{u}^{2}}-4u-5=0\\\\ \ \ \ \ \ (u-5)(u+1)=0\\\\ \ \therefore \ \ \ u=5\ \text{or}\ u=-1\\\\ \ \therefore \ \ \ {{5}^{{{{{\log }}_{{10}}}x}}}=5\ \text{or}\ {{5}^{{{{{\log }}_{{10}}}x}}}=-1\\\\ \ \ \ \ \ \text{Since}\ {{5}^{{{{{\log }}_{{10}}}x}}}>0,\ {{5}^{{{{{\log }}_{{10}}}x}}}=-1\ \text{is}\ \text{impossible}\text{.}\\\\ \ \therefore \ \ \ {{5}^{{{{{\log }}_{{10}}}x}}}=5\\\\ \ \therefore \ \ \ {{\log }_{{10}}}x=1\\\\ \ \therefore \ \ \ x=10\\\\\\ \text{(c)}\ \ {{9}^{{{{{\log }}_{3}}\left( {{{{\log }}_{2}}x} \right)}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{{{{{\log }}_{3}}9}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{{{{{\log }}_{3}}{{3}^{2}}}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{2}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\ \ \ \ \ \ 2{{\left( {{{{\log }}_{2}}x} \right)}^{2}}-{{\log }_{2}}x-1=0\\\\ \ \ \ \ \ \left( {2\cdot {{{\log }}_{2}}x+1} \right)\left( {{{{\log }}_{2}}x-1} \right)=0\\\\ \ \ \ \ \ {{\log }_{2}}x=-\displaystyle\frac{1}{2}\ \text{or}\ {{\log }_{2}}x=1\\\\ \ \therefore \ \ \ x={{2}^{{-\frac{1}{2}}}}\ \text{or}\ x=2\\\\ \ \therefore \ \ \ x=\displaystyle\frac{1}{{\sqrt{2}}}\ \text{or}\ x=2\\\\\\ \text{(d)}\ \ {{\log }_{2}}x+{{\log }_{4}}x+{{\log }_{{16}}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x+{{\log }_{{{{2}^{2}}}}}x+{{\log }_{{{{2}^{4}}}}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x+\displaystyle\frac{1}{2}{{\log }_{2}}x+\displaystyle\frac{1}{4}{{\log }_{2}}x=7\\\\ \ \ \ \ \ \displaystyle\frac{7}{4}{{\log }_{2}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x=4\\\ \therefore \ \ \ x={{2}^{4}}=16\\\\\\ \text{(e)}\ \ {{\log }_{7}}2+{{\log }_{{49}}}x={{\log }_{{\displaystyle\frac{1}{7}}}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+{{\log }_{{{{7}^{2}}}}}x={{\log }_{{{{7}^{{-1}}}}}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+\displaystyle\frac{1}{2}{{\log }_{7}}x=-{{\log }_{7}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+{{\log }_{7}}{{x}^{{\frac{1}{2}}}}={{\log }_{7}}{{3}^{{-\frac{1}{2}}}}\\\\ \ \ \ \ \ {{\log }_{7}}2{{x}^{{\frac{1}{2}}}}={{\log }_{7}}\displaystyle\frac{1}{{{{3}^{{\frac{1}{2}}}}}}\\\\ \ \ \therefore \ \ 2{{x}^{{\frac{1}{2}}}}=\displaystyle\frac{1}{{{{3}^{{\frac{1}{2}}}}}}\\\\ \ \ \therefore \ \ 4x=\displaystyle\frac{1}{3}\\\ \ \therefore \ \ x=\displaystyle\frac{1}{{12}}\\\\\\ \text{(f)}\ \ {{\log }_{3}}x-{{\log }_{{\displaystyle\frac{1}{3}}}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}x-{{\log }_{{{{3}^{{-1}}}}}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}x+{{\log }_{3}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}{{x}^{3}}=6\\\\ \ \ \ \ \ {{x}^{3}}={{3}^{6}}\\\\ \ \therefore \ \ \ x={{3}^{2}}=9\\\\\\ \text{(g)}\ \ {{\log }_{x}}(9{{x}^{2}})\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\ \ \ \ \ \ \displaystyle\frac{{{{{\log }}_{3}}(9{{x}^{2}})}}{{{{{\log }}_{3}}x}}\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\ \ \ \ \ \ {{\log }_{3}}(9{{x}^{2}})\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ ({{\log }_{3}}9+{{\log }_{3}}{{x}^{2}})\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ ({{\log }_{3}}{{3}^{2}}+2{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ (2+2{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \therefore \ \ (1+{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=2\\\\ \ \therefore \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}+\left( {{{{\log }}_{3}}x} \right)-2=0\\\\ \ \therefore \ \ ({{\log }_{3}}x+2)({{\log }_{3}}x-1)=0\\\\ \ \therefore \ \ \ {{\log }_{3}}x=-2\ \text{or}\ {{\log }_{3}}x=1\\\\ \ \therefore \ \ \ x={{3}^{{-2}}}\ \text{or}\ x=3\\\\ \ \therefore \ \ \ x=\displaystyle\frac{1}{9}\ \text{or}\ x=3 \end{array}$


            Solve the following equations.

$\begin{array}{l} 1.\quad 3^{{2 x-3}}=27^{{2 x}} \\\\ 2.\quad 5^{{x^{2}-9}}=1 \\\\ 3.\quad 5^{{x+1}}=\displaystyle \frac{1}{625}\\\\ 4.\quad \left(\displaystyle \frac{1}{2}\right)^{x}=64 \\\\ 5.\quad 2^{{3 x}} \cdot 4^{{x+1}}=128 \\\\ 6.\quad 3^{{x+1}} \cdot 9^{2-x}=\displaystyle \frac{1}{27} \\\\ 7.\quad \displaystyle \frac{27^{{2 x}}}{3^{{5-x}}}=\displaystyle \frac{3^{{2 x+1}}}{9^{{x+3}}}\\\\ 8.\quad 8^{{x-1}}=\left(\displaystyle \frac{1}{32}\right)^{x+1}\\\\ 9.\quad 10^{{-x}}=0.000001 \\\\ 10.\quad 4^{{x}}+4^{{x+1}}=20 \\\\ 11.\quad 4 \cdot 2^{{2 x}}+3 \cdot 2^{{x}}-1=0 \end{array}$

Show/Hide Solution

$\begin{array}{l}1.\quad\quad {{3}^{{2x-3}}}={{27}^{{2x}}}\\\ \ \ \ {{3}^{{2x-3}}}={{({{3}^{3}})}^{{2x}}}\\\ \ \ \ {{3}^{{2x-3}}}={{3}^{{6x}}}\\\ \therefore \ \ 2x-3=6x\\\ \therefore \ \ 8x=3\\\ \therefore \ \ x=\displaystyle \frac{3}{8}\\\\2.\quad\quad {{5}^{{{{x}^{2}}-9}}}=1\\\ \ \ \ {{5}^{{{{x}^{2}}-9}}}={{5}^{0}}\\\ \therefore \ \ {{x}^{2}}-9=0\\\ \ \ \ {{x}^{2}}=9\\\ \therefore \ \ x=\pm 3\\\\3.\quad\quad {{5}^{{x+1}}}=\displaystyle \frac{1}{{625}}\\\ \ \ \ {{5}^{{x+1}}}=\displaystyle \frac{1}{{{{5}^{4}}}}\\\ \ \ \ {{5}^{{x+1}}}={{5}^{{-4}}}\\\ \therefore \ \ x+1=-4\\\ \therefore \ \ x=-5\\\\4.\quad\quad {{\left( {\displaystyle \frac{1}{2}} \right)}^{x}}=64\\\ \ \ \ {{2}^{{-x}}}={{2}^{6}}\\\ \therefore \ \ -x=6\\\ \therefore \ \ x=6\\\\5.\quad\quad {{2}^{{3x}}}\cdot {{4}^{{x+1}}}=128\\\ \ \ \ {{2}^{{3x}}}\cdot {{\left( {{{2}^{2}}} \right)}^{{x+1}}}={{2}^{7}}\\\ \ \ \ {{2}^{{5x+2}}}={{2}^{7}}\\\ \therefore \ \ 5x+2=7\\\ \therefore \ \ 5x=5\\\ \therefore \ \ x=1\\\\6.\quad\quad {{3}^{{x+1}}}\cdot {{9}^{{2-x}}}=\displaystyle \frac{1}{{27}}\\\ \ \ \ {{3}^{{x+1}}}\cdot {{\left( {{{3}^{2}}} \right)}^{{2-x}}}=\displaystyle \frac{1}{{{{3}^{3}}}}\\\ \ \ \ {{3}^{{5-x}}}={{3}^{{-3}}}\\\ \therefore \ \ 5-x=-3\\\ \therefore \ \ x=8\\\ \therefore \ \ x=1\\\\7.\quad\quad \displaystyle \frac{{{{{27}}^{{2x}}}}}{{{{3}^{{5-x}}}}}=\displaystyle \frac{{{{3}^{{2x+1}}}}}{{{{9}^{{x+3}}}}}\\\ \ \ \ \displaystyle \frac{{{{{\left( {{{3}^{3}}} \right)}}^{{2x}}}}}{{{{3}^{{5-x}}}}}=\displaystyle \frac{{{{3}^{{2x+1}}}}}{{{{{\left( {{{3}^{2}}} \right)}}^{{x+3}}}}}\\\ \ \ \ {{3}^{{7x-5}}}={{3}^{{-5}}}\\\ \therefore \ \ 7x-5=-5\\\ \therefore \ \ x=0\\\ \therefore \ \ x=1\\\\8.\quad\quad {{8}^{{x-1}}}={{\left( {\displaystyle \frac{1}{{32}}} \right)}^{{x+1}}}\\\ \ \ \ {{\left( {{{2}^{3}}} \right)}^{{x-1}}}={{\left( {\displaystyle \frac{1}{{{{2}^{5}}}}} \right)}^{{x+1}}}\\\ \ \ \ {{2}^{{3x-3}}}={{2}^{{-5x-5}}}\\\ \therefore \ \ 3x-3=-5x-5\\\ \therefore \ \ 8x=8\\\ \therefore \ \ x=1\\\\9.\quad\quad {{10}^{{-x}}}=0.000001\\\ \ \ \ {{10}^{{-x}}}=\displaystyle \frac{1}{{1000000}}\\\ \ \ \ {{10}^{{-x}}}=\displaystyle \frac{1}{{{{{10}}^{6}}}}\\\ \ \ \ {{10}^{{-x}}}={{10}^{{-6}}}\\\ \therefore \ \ -x=-6\\\ \therefore \ \ x=6\\\\10.\;\ {{10}^{{-x}}}=0.000001\\\ \ \ \ {{10}^{{-x}}}=\displaystyle \frac{1}{{1000000}}\\\ \ \ \ {{10}^{{-x}}}=\displaystyle \frac{1}{{{{{10}}^{6}}}}\\\ \ \ \ {{10}^{{-x}}}={{10}^{{-6}}}\\\ \therefore \ \ -x=-6\\\ \therefore \ \ x=6\\\\11.\;\ 4\cdot {{2}^{{2x}}}+3\cdot {{2}^{x}}-1=0\\\ \ \ \ 4\cdot {{\left( {{{2}^{x}}} \right)}^{2}}+3\cdot {{2}^{x}}-1=0\\\ \ \ \ \text{Let}\ {{2}^{x}}=a,\ \text{then we have}\\\ \therefore \ \ 4{{a}^{2}}+3a-1=0\\\ \therefore \ \ (4a-1)(a+1)=0\\\ \therefore \ \ a=\displaystyle \frac{1}{4}\ \text{or}\ a=-1\\\ \therefore \ \ {{2}^{x}}=\displaystyle \frac{1}{4}\ \text{or}\ {{2}^{x}}=-1\ (\text{impossible})\\\ \ \ \ \text{Since}\ {{2}^{x}}>0,\ {{2}^{x}}=-1\ \text{is impossible}\text{.}\\\ \therefore \ \ {{2}^{x}}=\displaystyle \frac{1}{4}=\displaystyle \frac{1}{{{{2}^{2}}}}={{2}^{{-2}}}\\\therefore \ \ x=-2\end{array}$


1.         Write the following in radical form.

$\begin{array}{ll} \text{(a)}\quad (5)^{^{\tfrac{1}{2}}} & \text{(b)}\quad (-9)^{^{\tfrac{1}{3}}} \\\\ \text{(c)}\quad (2)^{^{-\frac{1}{2}}} & \text{(d)}\quad \left(-\displaystyle\frac{3}{4}\right)^{^{\tfrac{2}{5}}}\\\\ \text{(e)}\quad \left(\displaystyle\frac{2}{7}\right)^{^{\tfrac{5}{2}}} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ \ {{\left( 5 \right)}^{{\tfrac{{1\,}}{{2\,}}}}}=\sqrt{5}\\\\ \text{(b)}\ \ {{\left( {-9} \right)}^{{\tfrac{1}{3}}}}=\sqrt[3]{{-9}}\\\\ \text{(c)}\ \ {{\left( 2 \right)}^{{-\tfrac{{1\,}}{{2\,}}}}}=\displaystyle \frac{1}{{{{{\left( 2 \right)}}^{{\tfrac{{1\,}}{{2\,}}}}}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\\ \text{(d)}\ \ {{\left( {-\displaystyle \frac{3}{4}} \right)}^{{\tfrac{2}{5}}}}=\sqrt[5]{{{{{\left( {-\displaystyle \frac{3}{4}} \right)}}^{2}}}}=\sqrt[5]{{\displaystyle \frac{9}{{16}}}}\\\\ \text{(e)}\ \ {{\left( {\displaystyle \frac{2}{7}} \right)}^{{\tfrac{5}{2}}}}=\sqrt{{{{{\left( {\displaystyle \frac{2}{7}} \right)}}^{5}}}}=\sqrt{{\displaystyle \frac{{32}}{{16807}}}}\\ \end{array}$

2.         Write the following in fractional exponent form.

$\begin{array}{ll} \text{(a)}\quad \sqrt[6]{c^{^{5}}} & \text{(b)}\quad \sqrt[3]{-2} \\\\ \text{(c)}\quad \sqrt[5]{a^{^{4}} \sqrt[3]{b^{^{5}}}} & \text{(d)}\quad \sqrt[4]{\left(\displaystyle\frac{3}{7}\right)^{^{3}}} \end{array}$

Show/Hide Solution

$\begin{array}{l} (\text{a})\ \ \sqrt[6]{{{{c}^{5}}}}={{c}^{{\frac{5}{6}}}}\\\\ (\text{b})\ \ \sqrt[3]{{-2}}={{(-2)}^{{\frac{1}{3}}}}\\\\ (\text{c})\ \ \sqrt[5]{{{{a}^{4}}\sqrt[3]{{{{b}^{5}}}}}}=\sqrt[5]{{{{a}^{4}}{{b}^{{\frac{5}{3}}}}}}={{\left( {{{a}^{4}}{{b}^{{\frac{5}{3}}}}} \right)}^{{\frac{1}{5}}}}={{a}^{{\frac{4}{5}}}}{{b}^{{\frac{1}{3}}}}\\\\ (\text{d})\ \ \sqrt[4]{{{{{\left( {\displaystyle \frac{3}{7}} \right)}}^{3}}}}={{\left( {\displaystyle \frac{3}{7}} \right)}^{{\frac{3}{4}}}} \end{array}$

3.         Change the expression with the same radical and simplify the radicands.

$\begin{array}{ll} \text{(a)}\quad 6 \sqrt{2} \quad& \text{(b)}\quad 3 a \sqrt[3]{x} \\\\ \text{(c)}\quad 2 \sqrt[5]{2} \quad& \text{(d)}\quad \sqrt[4]{\displaystyle\frac{1}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{^{3}}} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad 6 \sqrt{2}=\sqrt{6^{2} \cdot 2}=\sqrt{72}\\\\ \text{(b)}\quad 3 a \sqrt[3]{x}=\sqrt[3]{3^{3} \cdot a^{3} x}=\sqrt[3]{27 a^{3} x}\\\\ \text{(c)}\quad 2 \sqrt[5]{2}=\sqrt[5]{2^{5} \cdot 2}=\sqrt[5]{64}\\\\ \text{(d)}\quad 3 \sqrt[4]{\displaystyle\frac{1}{2}}=\sqrt[4]{3^{4} \cdot \displaystyle\frac{1}{2}}=\sqrt[4]{\displaystyle\frac{81}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{3}}=\sqrt{3^{2} \cdot x^{3}}=\sqrt{9 x^{3}} \end{array}$

4.         Simplify.

$\begin{array}{ll} \text{(a)}\quad \sqrt{32} & \text{(b)}\quad \sqrt[5]{-32} \\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{^{16}}}{16 y^{^{4}}}} & \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{^{2}}}{4 y}} \\\\ \text{(e)}\ \displaystyle\frac{9^{^{\tfrac{1}{2}}}}{\sqrt[3]{27}}& \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{^{3}}}} & \text{(h)}\quad \sqrt[n]{\displaystyle\frac{32}{2^{^{5+n}}}} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad \sqrt{32}=\sqrt{16 \cdot 2}=4 \sqrt{2}\\\\ \text{(b)}\quad \sqrt[5]{-32}=\sqrt[5]{(-2)^{5}}=-2\\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{16}}{16 y^{4}}}=\sqrt[4]{\displaystyle\frac{3^{4}\left(x^{4}\right)^{4}}{2^{4} \cdot y^{4}}}=\displaystyle\frac{3 x^{4}}{2 y}\\\\ \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{2}}{4 y}}=\sqrt[3]{\displaystyle\frac{3^{3} \cdot 3 x^{2}}{4 y}}=3 \sqrt[3]{\displaystyle\frac{3 x^{2}}{4 y}}\\\\ \text{(e)}\quad \displaystyle\frac{9^{\displaystyle\frac{1}{2}}}{\sqrt[3]{27}}=\displaystyle\frac{\left(3^{2}\right)^{\frac{1}{2}}}{\sqrt[3]{3^{3}}}=\displaystyle\frac{3}{3}=1\\\\ \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{2}{3} \times \displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{25}{49}}=\displaystyle\frac{5}{7}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{3}}}=\sqrt[3]{\displaystyle\frac{(-6)^{3}}{2^{3} \times 10^{3}}}=\displaystyle\frac{-6}{2 \times 10}=-\displaystyle\frac{3}{10}\\\\ \text{(h)}\quad \sqrt[4]{\displaystyle\frac{32}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{2^{5}}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{1}{2^{n}}}=\displaystyle\frac{1}{2} \end{array}$

5.         Rationalize the denominators.

$\begin{array}{ll} \text{(a)}\quad \displaystyle\frac{4 \sqrt{35}}{3 \sqrt{7}}\quad & \text{(b)}\quad \displaystyle\frac{20}{\sqrt{5}} \\\\ \text{(c)}\quad \displaystyle\frac{18}{\sqrt[3]{2}}\quad& \text{(d)}\quad \displaystyle\frac{\sqrt[3]{32}}{\sqrt[4]{27}} \\\\ \text{(e)}\quad \displaystyle\frac{\sqrt[3]{36 a^{^{2}}}}{\sqrt[3]{9 a}}\quad & \text{(f)}\quad \displaystyle\frac{\sqrt[3]{2}}{\sqrt[6]{12}} \\\\ \text{(g)}\quad \displaystyle\frac{1}{\sqrt[3]{x y^{^{2}}}}\quad & \text{(h)}\quad \sqrt[m]{{\displaystyle\frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}} \end{array}$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \frac{{4\sqrt{{35}}}}{{3\sqrt{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{35}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{7\times 5}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{{4\sqrt{5}}}{3}\\\\\\\text{(b)}\ \ \ \ \displaystyle \frac{{20}}{{\sqrt{5}}}\\\\\ \ \ \ \ =\displaystyle \frac{{20}}{{\sqrt{5}}}\times \displaystyle \frac{{\sqrt{5}}}{{\sqrt{5}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{20\sqrt{5}}}{5}\\\\\ \ \ \ \ =4\sqrt{5}\\\\\\\text{(c)}\ \ \ \ \displaystyle \frac{{18}}{{\sqrt[3]{2}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18}}{{\sqrt[3]{2}}}\times \displaystyle \frac{{\sqrt[3]{{{{2}^{2}}}}}}{{\sqrt[3]{{{{2}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{{\sqrt[3]{{{{2}^{3}}}}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{2}\\\ \ \ \ \ =9\sqrt[3]{4}\\\\\text{(d)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{32}}}}{{\sqrt[4]{{27}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{2}^{3}}\cdot 4}}}}{{\sqrt[4]{{{{3}^{3}}}}}}\times \displaystyle \frac{{\sqrt[4]{3}}}{{\sqrt[4]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{2\sqrt[3]{4}\sqrt[4]{3}}}{3}\\\\\\\text{(e)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{36{{a}^{2}}}}}}{{\sqrt[3]{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{\displaystyle \frac{{36{{a}^{2}}}}{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{4a}}\\\\\\\text{(f)}\ \ \ \ \ \displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{12}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{{{2}^{2}}}}\cdot \sqrt[6]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[3]{2}\cdot \sqrt[6]{3}}}\times \displaystyle \frac{{\sqrt[6]{{{{3}^{5}}}}}}{{\sqrt[6]{{{{3}^{5}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[6]{{243}}}}{3}\\\\\\\text{(g)}\ \ \ \ \ \displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\times \displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{2}}y}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{3}}{{y}^{3}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{xy}}\\\\\text{(h)}\ \ \ \ \ \sqrt[m]{{\displaystyle \frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}}\\\\\ \ \ \ \ =\sqrt[m]{{\displaystyle \frac{{2y}}{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\ \displaystyle \frac{{\sqrt[m]{{2y}}}}{{\sqrt[m]{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2y}}}}{{y\sqrt[m]{{{{3}^{2}}{{x}^{3}}}}}}\times \displaystyle \frac{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2\cdot {{3}^{{m-2}}}{{x}^{{m-3}}}y}}}}{{3xy}}\end{array}$

6.         Reduce the order as far as possible.

$\begin{array}{ll} \text{(a)}\quad \sqrt[4]{25} \quad& \text{(b)}\quad \sqrt[6]{4} \\\\ \text{(c)}\quad \sqrt[6]{8} \quad& \text{(d)}\quad \sqrt[9]{8 y^{^{3}}} \\\\ \text{(e)}\quad \sqrt[6]{27^{^{3}}} \quad& \text{(f)}\quad \sqrt[8]{a^{2} b^{^{4}}} \\\\ \text{(g)}\quad \sqrt[12]{64 a^{^{2}} b^{^{6}}} & \text{(h)}\quad (72)^{^{\tfrac{3}{5}}} \\\\ \text{(i)}\quad \sqrt[3]{768} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad \sqrt[4]{25}=\sqrt[4]{5^{2}}=\sqrt{5}\\\\ \text{(b)}\quad \sqrt[6]{4}=\sqrt[6]{2^{2}}=\sqrt[3]{2}\\\\ \text{(c)}\quad \sqrt[6]{8}=\sqrt[6]{2^{3}}=\sqrt{2}\\\\ \text{(d)}\quad \sqrt[9]{8 y^{3}}=\sqrt[9]{2^{3} \cdot y^{3}}=\sqrt[3]{2 y}\\\\ \text{(e)}\quad \sqrt[6]{27^{3}}=\sqrt{27}=3 \sqrt{3}\\\\ \text{(f)}\quad \sqrt[8]{a^{2} b^{4}}=\sqrt[4]{a b^{2}}\\\\ \text{(g)}\quad \sqrt[12]{64 a^{2} b^{6}}=\sqrt[12]{8^{2} \cdot a^{2} b^{6}}=\sqrt[6]{8 a b^{3}}\\\\ \text{(h)}\quad (72)^{\frac{3}{5}}=\sqrt[5]{72^{3}}=\sqrt[5]{\left(3^{2} \cdot 2^{3}\right)^{3}}=\sqrt[5]{3^{6} \cdot 2^{9}}=6 \sqrt[5]{3 \cdot 16}=6 \sqrt[5]{48}\\\\ \text{(i)}\quad \sqrt[3]{768}=\sqrt[3]{4^{3} \cdot 12}=4 \sqrt[3]{12} \end{array}$

7.         Find the simplified forms.

$\begin{array}{ll} \text{(a)}\quad \sqrt{\displaystyle\frac{9}{50}} \quad& \text{(b)}\quad \sqrt[3]{\displaystyle\frac{-192}{49}} \\\\ \text{(c)}\quad \sqrt[4]{16}\quad& \text{(d)}\quad 2 \sqrt[3]{56} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad\sqrt{\displaystyle \frac{9}{50}}=\sqrt{\displaystyle \frac{3^{2}}{5^{2} \cdot 2}}=\displaystyle \frac{3}{5 \sqrt{2}} \times \displaystyle \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{3 \sqrt{2}}{10}\\\\ \text{(b)}\quad\sqrt[3]{\displaystyle \frac{-192}{49}}=\sqrt[3]{\displaystyle \frac{(-4)^{3} \cdot 3}{7^{2}} \times \displaystyle \frac{7}{7}}=-\displaystyle \frac{4 \sqrt[3]{21}}{7}\\\\ \text{(c)}\quad\sqrt[4]{16}=\sqrt[4]{2^{4}}=2\\\\ \text{(d)}\quad 2 \sqrt[3]{56}=2 \sqrt[3]{2^{3} \cdot 7}=4 \sqrt[3]{7} \end{array}$

1.         Sketch the following lines.

$\begin{array} \text{(a)}\ y=3\\\\ \text{(b)}\ x=-2\\\\ \text{(c)}\ y=5x\\\\ \text{(d)}\ y=-3x\\\\ \text{(e)}\ y=\displaystyle\frac{1}{2}x \end{array}$

Show/Hide Solution

$\text{(a)}\ y=3$


$\text{(b)}\ x=-2$


$\begin{array}{l} \text{(c)}\ y=5x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=5x & 0 & 5\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(d)}\ y=-3x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x & 0 & -3\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(e)}\ y=\displaystyle\frac{1}{2}x\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x & 0 & -3\\ \hline \end{array}\end{array}$


2.         Graph each line in the same coordinate plane.

$\begin{array}{lll} \text{(a)}\ y=x+2 \\\\ \text{(b)}\ y=x+3 \\\\ \text{(c)}\ y=x+5 \\\\ \text{(d)}\ y=x-1 \\\\ \text{(e)}\ y=x-2 \\\\ \text{(f)}\ y=x-4 \end{array}$

Show/Hide Solution


$\bullet\quad y=x$ is a line with slope equal to 1 and passing through origin.

$\bullet\quad y=x + 2$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 2).

$\bullet\quad y=x + 3$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 3).

$\bullet\quad y=x + 5$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, 5).

$\bullet\quad y=x -1 $ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -1).

$\bullet\quad y=x -2$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -2).

$\bullet\quad y=x -4$ is a line which is parallel to $y=x$ and cutting $y$-axis at (0, -4).


3.         Graph each line.

$\begin{array} \text{(a)}\ y=2x+2\\\\ \text{(b)}\ y=\displaystyle\frac{1}{2}x+2\\\\ \text{(c)}\ y=-\displaystyle\frac{1}{2}x+2 \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ y=2x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & -1\\ \hline y=2x+2 & 2 & 0\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(b)}\ y=\displaystyle\frac{1}{2}x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & -4\\ \hline y=\displaystyle\frac{1}{2}x+2 & 2 & 0\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(b)}\ y=-\displaystyle\frac{1}{2}x+2\\ \begin{array}{|l|l|l|} \hline x & 0 & 4\\ \hline -\displaystyle\frac{1}{2}x+2 & 2 & 0\\ \hline \end{array}\end{array}$



4.         Find the slope and $y$-intercept for the following equations and sketch their graphs.

$\begin{array}{l} \text{(a)}\ y=x-2 \\\\ \text{(b)}\ y=-3x-3 \\\\ \text{(c)}\ y=\displaystyle\frac{1}{2}x+1\\\\ \text{(d)}\ y=-\displaystyle\frac{1}{2}x+1 \\\\ \text{(e)}\ y+3x=3 \\\\ \text{(f)}\ x-y=3 \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ y=x-2\\\\ \text{slope}= 1\\\\ y\text{-intercept}=-2\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 2\\ \hline y=x-2 & -2 & 0\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(b)}\ y=-3x-3\\\\ \text{slope}= -3\\\\ y\text{-intercept}=-3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x-3 & -3 & 0\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(c)}\ y=\displaystyle\frac{1}{2}x+1\\\\ \text{slope}= \displaystyle\frac{1}{2}\\\\ y\text{-intercept}=1\\\\ \begin{array}{|l|l|l|} \hline x & 0 & -2\\ \hline y=\displaystyle\frac{1}{2}x+1 & 1 & 0\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(d)}\ y=-\displaystyle\frac{1}{2}x+1\\\\ \text{slope}= -\displaystyle\frac{1}{2}\\\\ y\text{-intercept}=1\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 2\\ \hline y=-\displaystyle\frac{1}{2}x+1 & 1 & 0\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(e)}\ y+3x=3\\\\ \quad y=-3x+3\\\\ \text{slope}=-3\\\\ y\text{-intercept}=3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 1\\ \hline y=-3x+3 & 3 & 0\\ \hline \end{array}\end{array}$


$\begin{array}{l} \text{(f)}\ x-y=3\\\\ \quad y=x-3\\\\ \text{slope}=1\\\\ y\text{-intercept}=-3\\\\ \begin{array}{|l|l|l|} \hline x & 0 & 3\\ \hline y=x-3 & -3 & 0\\ \hline \end{array}\end{array}$



5.         Find the equation of the straight line with the given slope and $y$-intercept.

(a) slope 3, $y$-intercept 4

(b) slope 2, $y$-intercept 0

(c) slope 0, $y$-intercept 2

(d) slope 0, $y$-intercept 0

Show/Hide Solution

$\begin{array}{l} \text{(a)}\quad m=3,\ c=4\\\\ \quad \quad \ \ y=m x+c\\\\ \quad \quad \ \ \ \ =3 x+4 \end{array}$

$\therefore$ The equation of the line is $y=3 x+4$.

$\begin{array}{l} \text{(b)}\quad m=2,\ c=0\\\\ \quad \quad \ \ y=m x+c\\\\ \quad \quad \ \ \ \ =2 x+0\\\\ \quad \quad \ \ \ \ =2 x \end{array}$

$\therefore$ The equation of the line is $y=2x$.

$\begin{array}{l} \text{(c)}\quad m=0,\ c=2\\\\ \quad \quad \ \ y=(0) x+2\\\\ \quad \quad \ \ \ \ =2 \end{array}$

$\therefore$ The equation of the line is $y=2$.

$\begin{array}{l} \text{(d)}\quad m=0,\ c=0\\\\ \quad \quad \ \ y=(0) x+0\\\\ \quad \quad \ \ \ \ =0 \end{array}$

$\therefore$ The equation of the line is $y=0$.


6.         Find the equation of the line which has a slope $m$ of $\displaystyle -\frac{2}{3}$ and passes through the point $(9,4)$.

Show/Hide Solution

Let $\left(x_{1}, y_{1}\right)=(9,4). m=-\displaystyle \frac{2}{3}$

$y-y_{1}=m\left(x-x_{1}\right)$

$y-4=-\displaystyle \frac{2}{3}(x-9)=-\displaystyle \frac{2}{3} x+6$

$y=-\displaystyle \frac{2}{3} x+10$

$\therefore 2x+3y=30$

$\therefore$ The equation of the line is $2x+3y=30$.

7.         A line has slope $-2$ and $y$-intercept $6$, find its $x$-intercept.

Show/Hide Solution

$$\begin{array}{l} m=-2,\ c=6 \\\\ \text{The equation of the line having slope} \\\\ m\ \text{and}\ y\text{-intercept}\ c\ \text{is}\ y=m x+c.\\\\ \therefore\ \text{The equation of given line is}\ y=-2x + 6.\\\\ \text { When }\ y=0,-2 x+6=0\Rightarrow x=3 \\\\ \therefore x \text {-intercept }=3 \end{array}$$

8.         Find the equation of the line which:

(a) has a slope of $5$ and passes through the point $(2,9)$.

(b) has a slope of $1$ and passes through the point $(1,-2)$.

(c) has a slope of $-3$ and passes through the point $(-1,6)$.

(d) has a slope of $-2$ and passes through the point $(-1,4)$.

Show/Hide Solution

$$\begin{array}{l} \text { (a) } \text{Let}\ \left(x_{1}, y_{1}\right)=(2,9).\ m=5\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-9 =5(x-2)\\\\ \quad \quad y-9 =5 x-10 \\\\ \quad \quad y=5 x-1 \end{array}$$

$$\begin{array}{l} \text { (b) } \text{Let}\ \left(x_{1}, y_{1}\right)=(1,-2). \ m=1\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-(-2)=1(x-1)\\\\ \quad \quad y+2 =x-1 \\\\ \quad \quad y=x-3 \end{array}$$

$$\begin{array}{l} \text { (c) } \text{Let}\ \left(x_{1}, y_{1}\right)=(-1,6).\ m=-3\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-6=-3\left(x-(-1)\right)\\\\ \quad \quad y-6 =-3 x-3 \\\\ \quad \quad y=-3 x+3 \end{array}$$

$$\begin{array}{l} \text { (d) } \text{Let}\ \left(x_{1}, y_{1}\right)=(-1,4).\ m=-2\\\\ \quad \quad \text{The equation of the line having slope} \\\\ \quad \quad m\ \text{passing through the point}\ \left(x_{1}, y_{1}\right)\text{is}\\\\ \quad \quad y-y_{1}=m\left(x-x_{1}\right)\\\\ \quad \quad \text{Hence, the required line is}\\\\ \quad \quad y-4=-2\left(x-(-1)\right) \\\\ \quad \quad y-4 =-2 x-2 \\\\ \quad \quad y=-2 x+2 \end{array}$$


9.         Find the slope and equation of the line joining the following pairs of points.

(a) $(2,4)$ and $(6,8)$

(b) $(-3,5)$ and $(6,-1)$

(c) $(-2, 1)$ and $(-4, -2)$

Show/Hide Solution

(a) Let $\left(x_{1}, y_{1}\right)=(2,4)$ and $\left(x_{2}, y_{2}\right)=(6,8)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{8-4}{6-2}=\frac{4}{4}=1$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-4=1(x-2)$

$y-4=x-2$

$y=x+2$

(b) Let $\left(x_{1}, y_{1}\right)=(-3,5)$ and $\left(x_{2}, y_{2}\right)=(6,-1)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{-1-5}{6-(-3)}=-\frac{6}{9}=-\frac{2}{3}$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-5=-\displaystyle\frac{2}{3}\left(x-(-3)\right)$

$2x+3y = 9$

(c) Let $\left(x_{1}, y_{1}\right)=(-2,1)$ and $\left(x_{2}, y_{2}\right)=(-4,-2)$.

The slope of the line passing through the points $\left(x_{1}, y_{1}\right) and \left(x_{2}, y_{2}\right)$ is

$m=\displaystyle\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\therefore\ m=\displaystyle\frac{-2-1}{-4-(-2)}=\frac{3}{2}$

The equation of the line having slope $m$ and passing through the point $\left(x_{1}, y_{1}\right)$ is

$y-y_{1}=m\left(x-x_{1}\right)$.

Hence, the equation of the required line is

$y-1=\displaystyle\frac{3}{2}\left(x-(-2)\right)$

$3x-2y = -8$


10.        Determine which of the pairs of lines in each case with given equations are parallel or perpendicular or neither.

(a) $y=3x-2$ and $y=3x+9$

(b) $y=\displaystyle \frac{2}{3}x-5$ and $y=\displaystyle \frac{3}{2}x-5$

(c) $y=3x-2$ and $y=-\displaystyle \frac{1}{3}x+9$

(d) $y=\displaystyle \frac{2}{3}x-5$ and $y=-\displaystyle \frac{3}{2}x-5$

Show/Hide Solution

$\begin{array}{l}(\text{a})\ \ \ \ {{l}_{1}}:y=3x-2\ \text{and}\ {{l}_{2}}:y=3x+9\\\ \ \ \therefore \ \ {{m}_{1}}=3\ \text{and}\ {{m}_{2}}=3\\\ \ \ \therefore \ \ {{m}_{1}}={{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\parallel \ {{l}_{2}}\\\\(\text{b})\ \ \ \ {{l}_{1}}:y=\displaystyle \frac{2}{3}x-5\ \text{and}\ {{l}_{2}}:y=\displaystyle \frac{3}{2}x-5\\\ \ \ \therefore \ \ {{m}_{1}}=\displaystyle \frac{2}{3}\ \text{and}\ {{m}_{2}}=\displaystyle \frac{3}{2}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \ \ \ \ \text{Again,}\ {{m}_{1}}\cdot {{m}_{2}}=\displaystyle \frac{2}{3}\times \displaystyle \frac{3}{2}=1\ne -1\\\ \ \ \therefore \ \ {{l}_{1}}\ \text{and}\ {{l}_{2}}\ \text{are neither parallel nor }\\\ \ \ \ \ \ \text{perpendicular to each other}\text{.}\\\\(\text{c})\ \ \ \ {{l}_{1}}:y=3x-2\ \text{and}\ {{l}_{2}}:y=-\displaystyle \frac{1}{3}x+9\\\ \ \ \therefore \ \ {{m}_{1}}=3\ \text{and}\ {{m}_{2}}=-\displaystyle \frac{1}{3}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\nparallel \ {{l}_{2}}\\\ \ \ \ \ \ \text{But,}\ {{m}_{1}}\cdot {{m}_{2}}=3\left( {-\displaystyle \frac{1}{3}} \right)=-1\\\ \ \ \therefore \ \ {{l}_{1}}\bot \ {{l}_{2}}\\\\(\text{d})\ \ \ \ {{l}_{1}}:y=\displaystyle \frac{2}{3}x-5\ \text{and}\ {{l}_{2}}:y=-\displaystyle \frac{3}{2}x-5\\\ \ \ \therefore \ \ {{m}_{1}}=\displaystyle \frac{2}{3}\ \text{and}\ {{m}_{2}}=-\displaystyle \frac{3}{2}\\\ \ \ \therefore \ \ {{m}_{1}}\ne {{m}_{2}}\\\ \ \ \therefore \ \ {{l}_{1}}\nparallel \ {{l}_{2}}\\\ \ \ \ \ \ \text{But,}\ {{m}_{1}}\cdot {{m}_{2}}=\displaystyle \frac{2}{3}\left( {-\displaystyle \frac{3}{2}} \right)=-1\\\ \ \ \therefore \ \ {{l}_{1}}\bot \ {{l}_{2}}\end{array}$

11.         Find the equation of the line which is parallel to the line:

(a) with equation $y=4x+2$ and passes through $(0,8)$.

(b) with equation $y=-x+3$ and passes through ($0,5)$.

(c) with equation $y=-2x-3$ and passes through $(0,-7)$.

(d) with equation $y=-\displaystyle \frac{4}{5}x-3$ and passes through $ \displaystyle \left(0,\frac{1}{2}\right)$.

Show/Hide Solution

$\begin{array}{l}(\text{a})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=4x+2\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=4\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=4\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,8)}\ \text{is}\\\\\ \ \ \ y-8=4(x-0)\\\\\ \ \ \ y=4x+8\\\\(\text{b})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-x+3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=-1.\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,5)}\ \text{is}\\\\\ \ \ \ y-5=-(x-0)\\\\\ \ \ \ y=-x+5\\\\(\text{c})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-2x-3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-2\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}},\ {{m}_{2}}=-2.\text{ }\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}-\text{7)}\ \text{is}\\\\\ \ \ \ y-(-7)=-2(x-0)\\\\\ \ \ \ y=-2x-7\\\\(\text{d})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-\displaystyle \frac{4}{5}x-3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-\displaystyle \frac{4}{5}\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\parallel \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{4}{5}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}\displaystyle \frac{1}{2}\text{)}\ \text{is}\\\\\ \ \ \ y-\displaystyle \frac{1}{2}=-\displaystyle \frac{4}{5}(x-0)\\\\\ \ \ \ y=-\displaystyle \frac{4}{5}x+\displaystyle \frac{1}{2}\end{array}$

12.         Find the equation of the line which is perpendicular to the line:

(a) with equation $y=5x-4$ and passes through $(0,7)$

(b) with equation $y=-x+7$ and passes through $(0,4)$

(c) with equation $y=-2x+3$ and passes through $(0,-4)$

(d) with equation $ \displaystyle y=x-\frac{3}{2}$ and passes through $ \displaystyle \left( {0,\frac{5}{4}} \right)$

Show/Hide Solution

$\begin{array}{l}(\text{a})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=5x-4\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=5\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=-\displaystyle \frac{1}{5}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,7)}\ \text{is}\\\\\ \ \ \ y-7=-\displaystyle \frac{1}{5}(x-0)\\\\\ \ \ \ y=-\displaystyle \frac{1}{5}x+7\\\\(\text{b})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-x+7\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=1\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,4)}\ \text{is}\\\\\ \ \ \ y-4=1(x-0)\\\\\ \ \ \ y=x+4\\\\(\text{c})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=-2x+3\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=-2\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=\displaystyle \frac{1}{2}\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point (0,}-\text{4)}\ \text{is}\\\\\ \ \ \ y-(-4)=\displaystyle \frac{1}{2}(x-0)\\\\\ \ \ \ y=\displaystyle \frac{1}{2}x-4\\\\(\text{d})\ \ \text{Given line}\ \left( {{{l}_{1}}} \right):\ y=x-\displaystyle \frac{3}{2}\\\ \ \ \ \ \therefore \ \ {{m}_{1}}=1\\\\\ \ \ \ \text{Let the required line be }{{l}_{2}}.\\\\\ \ \ \ \text{Since }{{l}_{2}}\bot \ {{l}_{1}}\text{,}\ {{m}_{2}}=-\displaystyle \frac{1}{{{{m}_{1}}}}=-1\\\\\ \ \ \ \text{The equation of }{{l}_{2}}\ \text{passing }\\\ \ \ \ \text{through the point }\left( {\text{0,}\displaystyle \frac{5}{4}} \right)\ \text{is}\\\\\ \ \ \ y-\displaystyle \frac{5}{4}=-1(x-0)\\\\\ \ \ \ y=-x+\displaystyle \frac{5}{4}\end{array}$

13.         Show that the line through $(3n,0)$ and $(0,7n)$ is parallel to the line through $(0,21n)$ and $(9n,0)$.

Show/Hide Solution

$\begin{array}{l}\text{Let }{{l}_{1}}\ \text{contain }(3n,\ 0)\ \text{and}\ (0,\ 7n)\ \\\text{and}\ {{l}_{2}}\ \text{contain }(0,\ 21n)\ \text{and}\ (9n,\ 0).\\\\\therefore \ {{m}_{1}}=\displaystyle \frac{{7n-0}}{{0-3n}}=\displaystyle \frac{{7n}}{{-3n}}=-\displaystyle \frac{7}{3}\\\\\therefore \ {{m}_{2}}=\displaystyle \frac{{0-21n}}{{9n-0}}=\displaystyle \frac{{-21n}}{{9n}}=-\displaystyle \frac{7}{3}\\\\\therefore \ {{m}_{1}}={{m}_{2}}\\\\\therefore \ {{l}_{1}}\parallel {{l}_{2}}\end{array}$

14.         Prove that the triangle whose vertices are $H(-12,1)$, $K(9,3)$ and $M(11,-18)$ is a right triangle.

Show/Hide Solution

$ \begin{array}{l}H=(-12,1),\ K=(9,3),\ M=(11,-18)\\\\{{m}_{{HK}}}=\displaystyle \frac{{3-1}}{{9-(-12)}}=\displaystyle \frac{2}{{21}}\\\\{{m}_{{KM}}}=\displaystyle \frac{{-18-3}}{{11-9}}=-\displaystyle \frac{{21}}{2}\\\\\therefore {{m}_{{HK}}}\cdot {{m}_{{KM}}}=\displaystyle \frac{2}{{21}}\left( {-\displaystyle \frac{{21}}{2}} \right)=-1\\\\\therefore \ HK\bot KM\\\\\therefore \ \triangle HKM\ \text{is a right triangle}\text{.}\end{array}$

15.        Given the points $P(1,2)$, $Q(5,-6)$ and $R(b,b)$, determine the value of $b$ so that angle $PQR$ is a right angle.

Show/Hide Solution

$ \begin{array}{l}P=(1,2),\ Q=(5,-6),\ R=(b,b)\\\\{{m}_{{PQ}}}=\displaystyle \frac{{-6-2}}{{5-1}}=-2\\\\{{m}_{{QR}}}=\displaystyle \frac{{b-(-6)}}{{b-5}}=\displaystyle \frac{{b+6}}{{b-5}}\\\\\text{Since}\ \angle PQR\text{ is a right angle,}\\\\\therefore {{m}_{{PQ}}}\cdot {{m}_{{QR}}}=-1\\\\\therefore \ -2\left( {\displaystyle \frac{{b+6}}{{b-5}}} \right)=-1\\\\\therefore \ 2b+12=b-5\\\\\therefore b=17\end{array}$

16.        A right-angled isosceles triangle has vertices at $(0,5)$, $(5,0)$ and $(-5,0)$. Find the equation of each of the three sides.

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \text{Let}\ A=(0,\ 5),\ B=(-5,\ 0)\ \text{and}\ C=(5,\ 0).\\\\\therefore \ \ {{m}_{{AB}}}=\displaystyle \frac{{0-5}}{{-5-0}}\ =1\\\\\therefore \ \ \text{The equation of the line through }A\ \text{and}\ B\ \text{is}\\\ \ \text{ }\ y-5=1(x-0)\Rightarrow y=x+5\\\\\,\ \ \ \ {{m}_{{AC}}}=\displaystyle \frac{{0-5}}{{5-0}}\ =-1\\\\\therefore \ \ \text{The equation of the line through }A\ \text{and}\ C\ \text{is}\\\ \ \text{ }\ y-5=-1(x-0)\Rightarrow y=-x+5\\\\\ \ \ \ \text{Since }B\ \text{and}\ C\ \text{lie on }x\text{-axis, the equation }\\\ \ \ \ \text{of the line through }B\ \text{and}\ C\ \text{is}\ y=0.\end{array}$

17.         Determine the slope of each side of the quadrilateral whose vertices are $A(5,6)$, $B(13,6)$, $C(11,2)$ and $D(1,2)$. Can you tell what kind of a quadrilateral it is?

Show/Hide Solution

$\begin{array}{l}A=(5,6),\ B=(13,6),\ C=(11,2),\ D=(1,2)\\\\\text{The slope of the line passing through }\\({{x}_{1}},\ {{y}_{1}})\ \text{and}\ ({{x}_{2}},\ {{y}_{2}})\ \text{is}\ m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}.\\\\\therefore \ \ {{m}_{{AB}}}=\displaystyle \frac{{6-6}}{{13-5}}=0.\\\\\,\ \ \ \ {{m}_{{BC}}}=\displaystyle \frac{{2-6}}{{11-13}}=2.\\\\\,\ \ \ \ {{m}_{{CD}}}=\displaystyle \frac{{2-2}}{{1-13}}=0.\\\\\,\ \ \ \ {{m}_{{AD}}}=\displaystyle \frac{{2-6}}{{1-5}}=1.\\\\\therefore \ \ {{m}_{{AB}}}={{m}_{{CD}}}\\\\\therefore \ \ AB\parallel CD\\\\\therefore \ \ ABCD\ \text{is a trapezium}\text{.}\end{array}$

18.         Given the points $D(-4,6)$, $E(1,1)$, and $F(4,6)$, find the slopes of $DE$ and $EF$. Are the points $D$, $E$ and $F$ collinear, explain why?

Show/Hide Solution

$\begin{array}{l}D=(-4,6),\ E=(1,1),\ F=(4,6)\\\\\text{The slope of the line passing through }\\({{x}_{1}},\ {{y}_{1}})\ \text{and}\ ({{x}_{2}},\ {{y}_{2}})\ \text{is}\ m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}.\\\\\therefore \ \ {{m}_{{DE}}}=\displaystyle \frac{{1-6}}{{1-(-4)}}=-\displaystyle \frac{5}{5}=-1\\\\\,\ \ \ \ {{m}_{{EF}}}=\displaystyle \frac{{6-1}}{{4-1}}=\displaystyle \frac{5}{3}\\\\\therefore \ \ {{m}_{{DE}}}\ne {{m}_{{EF}}}\\\\\therefore \ \ D,\ E\ \text{and}\ F\ \text{are not collinear}\text{.}\end{array}$

19.        Prove that the quadrilateral with vertices $A(-2,2)$, $B(2,-2)$, $C(4,2)$ and $D(2,4)$ is a trapezoid with perpendicular diagonals.

Show/Hide Solution

$ D=(-4,6),\ E=(1,1),\ F=(4,6)$

The slope of the line passing through $({{x}_{1}},\ {{y}_{1}})$ and $({{x}_{2}},\ {{y}_{2}})$ is $m=\frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$.

$\begin{aligned} \therefore m_{AB}&=\displaystyle \frac{-2-2}{2+2} = -1\\\\ m_{BC}&=\displaystyle \frac{2+2}{4-2} =2\\\\ m_{CD}&=\displaystyle \frac{4-2}{2-4} = -1\\\\ m_{AD}&=\displaystyle \frac{4-2}{2+2} = \displaystyle \frac{1}{2}\\\\ m_{AC}&=\displaystyle \frac{2-2}{4+2} = 0\\\\ m_{BD}&=\displaystyle \frac{4+2}{2-2} =\text{undefined} \end{aligned}$

Since $m_{AB}=m_{CD},\ AB\parallel CD.$

$m_{AC} = 0\Rightarrow AC$ is a horizontal line.

$m_{BD} =\text{undefined}\Rightarrow BD$ is a vertical line.

$\therefore AC\bot BD$.

Hence, the quadrilateral $ABCD$ is a trapezoid with perpendicular diagonals.

20.         Find the slopes of the six lines determined by the points $A(-5,4)$, $B(3,5)$, $C(7,-2)$, $D(-1,-3)$. Prove that $ABCD$ is a rhombus.

Show/Hide Solution

$A=(-5,4),\ B=(3,5),\ C=(7,-2),\ D=(-1,-3)$

The slope of the line passing through $({{x}_{1}},\ {{y}_{1}})$ and $({{x}_{2}},\ {{y}_{2}})$ is $m=\displaystyle \frac{{{{y}_{2}}-{{y}_{1}}}}{{{{x}_{2}}-{{x}_{1}}}}$.

$\begin{aligned} m_{AB}&=\displaystyle \frac{5-4}{3+5}=\displaystyle \frac{1}{8}\\\\ m_{BC}&=\displaystyle \frac{-2-5}{7-3} =-\displaystyle \frac{7}{4} \\\\ m_{CD}&=\displaystyle \frac{-3+2}{-1-7} = \displaystyle \frac{1}{8}\\\\ m_{AD}&=\displaystyle \frac{-3-4}{-1+5} = -\displaystyle \frac{7}{4}\\\\ m_{AC}&=\displaystyle \frac{-2-4}{7+5} = -\displaystyle \frac{1}{2}\\\\\ m_{BD}&=\displaystyle \frac{-3-5}{-1-3}=2 \end{aligned}$

$\therefore\ m_{AB}=m_{CD}\Rightarrow AB\parallel CD$

$\quad\ m_{BC}=m_{AD}\Rightarrow BC\parallel AD$

$\quad\ m_{AC}\cdot m_{BD}=-\displaystyle \frac{1}{2}\times 2 = -1\Rightarrow AC\bot BD$

Hence $ABCD$ is a rhombus.